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我有以下结构

create table organisation (
  id uuid primary key,
  organisation_name varchar not null
);

create table organisation_member (
  id uuid primary key,
  email varchar not null,
  name varchar not null,
  organisation uuid not null,
  uid uuid,
  foreign key (organisation) references organisation(id),
  foreign key (uid) references users(id)
);

create table organisation_member_invite (
  id uuid primary key,
  organisation_member uuid not null,
  uid uuid not null,
  foreign key (organisation_member) references organisation_member(id),
  foreign key (uid) references users(id)
)

使用带有 supabase 的 Postgrest，我想查询所有organisations链接到organisation_member_invite. 在 SQL 中，我只会做一个简单的左连接

select *, o.* from organisation_member_invite as omi
right join organisation_member as om on om.id = omi.organisation_member
left join organisation as o on o.id = om.organisation
where omi.user = <user_id>

使用 postgrest，这变成了类似的东西

...from('organisation_member_invite').select('''
*,
organisation_member(
  organisation(*)
)
''').eq('user', uid);

不幸的是，在打印错误变量的几个部分时，我得到以下输出。

打印（错误提示）

Try changing 'organisation' to one of the following: 'organisation!organisation_member_organisation_fkey', 'organisation!organisation_function', 'organisation!organisation_member_invitation'. Find the desired relationship in the 'details' key.

打印（错误。详细信息）

details: [
    {relationship: organisation_member_organisation_fkey[organisation][id], embedding: organisation_member with organisation, cardinality: many-to-one}, 
    {relationship: public.organisation_function[organisation_function_member_fkey][organisation_function_organisation_fkey], embedding: organisation_member with organisation, cardinality: many-to-many}, 
    {relationship: public.organisation_member_invitation[organisation_member_invitation_membership_fkey][organisation_member_invitation_organisation_fkey], embedding: organisation_member with organisation, cardinality: many-to-many}
]

打印（错误消息）

Could not embed because more than one relationship was found for 'organisation_member' and 'organisation'

. organisation除了in的外键之外，我没有任何其他关系organisation_member。我真的很喜欢这种表示法，但它也限制了我表达我真正想要的数据。

我究竟做错了什么？

4

0 回答 0